A) \[\frac{\omega \,(M-2\,m)}{(M+2m)}\]
B) \[\frac{\omega M}{(M+2m)}\]
C) \[\frac{\omega \,M}{(M+m)}\]
D) \[\frac{\omega \,(M+2\,m)}{M}\]
Correct Answer: B
Solution :
| Key Idea: Angular momentum remains conserved in the universe. |
| According to conservation of angular momentum |
| \[L=\] constant |
| or \[l\omega =\text{constant}\] |
| \[\therefore \] \[{{l}_{1}}{{\omega }_{1}}={{l}_{2}}{{\omega }_{2}}\] |
| Initial moment of inertia |
| \[{{l}_{1}}=M{{R}^{2}}\] |
| and angular velocity |
| \[{{\omega }_{1}}=\omega \] |
| Hence, \[{{l}_{1}}{{\omega }_{1}}=M{{R}^{2}}\omega \] |
| When two objects of mass m are attached to opposite ends of a diameter, the final readings are |
| \[{{l}_{2}}=M\,{{R}^{2}}+m\,{{R}^{2}}+m\,{{R}^{2}}\] |
| \[=(M+2m)\,{{R}^{2}}\] |
| So, \[{{l}_{2}}{{\omega }_{2}}=(M+2m){{R}^{2}}{{\omega }_{2}}\] ....(iii) |
| \[\therefore \] From Eqs. (i), (ii) and (iii) |
| \[M{{R}^{2}}\,\omega \,=\,(M+2m)\,{{R}^{2}}\,{{\omega }_{2}}\] |
| \[\Rightarrow \] \[{{\omega }_{2}}=\frac{\omega M}{M+2m}\] |
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